Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

Asked by Abhisek | 11 months ago |  100

1 Answer

Solution :-

Let us consider the equation of the required circle be 

(x – h)+ (y – k)2 = r2

We know that the circle passes through points (2,3) and (-1,1).

(2 – h)2+ (3 – k)2 =r2 ……………..(1)

(-1 – h)2+ (1– k)2 =r2 ………………(2)

Since, the centre (h, k) of the circle lies on line x – 3y – 11= 0,

h – 3k =11………………… (3)

From the equation (1) and (2), we obtain

(2 – h)2+ (3 – k)2 =(-1 – h)2 + (1 – k)2

4 – 4h + h2 +9 -6k +k2 = 1 + 2h +h2+1 – 2k + k2

4 – 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11……………. (4)

Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,

6h+ 4k – 6(h-3k) = 11 – 66

6h + 4k – 6h + 18k = 11 – 66

22 k = – 55

K = \( \dfrac{-5}{2}\)

Substitute this value of K in equation (4) to get,

6h + 4(\( \dfrac{-5}{2}\)) = 11

6h – 10 = 11

6h = 21

h = \( \dfrac{21}{6}\)

h = \( \dfrac{7}{2}\)

We obtain h = \( \dfrac{7}{2}\)and k =\( \dfrac{-5}{2}\)

On substituting the values of h and k in equation (1), we get

(2 – \( \dfrac{7}{2}\))2 + (3 + \( \dfrac{5}{2}\))2 = r2

\( (\dfrac{4-7}{2})^2+(\dfrac{6+5}{2})^2=r^2\)

\( (\dfrac{-3}{2})^2\) + \( (\dfrac{11}{2})^2\)= r2

\( \dfrac{9}{4}\) + \( \dfrac{121}{4}\) = r2

\( \dfrac{130}{4}\) = r2

The equation of the required circle is

(x – \( \dfrac{7}{2}\))2 + (y + \( \dfrac{5}{2}\))2 = \( \dfrac{130}{4}\)

\( (\dfrac{2x-7}{2})^2+(\dfrac{2y+5}{2})^2=\dfrac{130}{4}\)

4x2 -28x + 49 +4y2 + 20y + 25 =130

4x2 +4y2 -28x + 20y – 56 = 0

4(x2 +y2 -7x + 5y – 14) = 0

x+ y– 7x + 5y – 14 = 0

The equation of the required circle is x+ y– 7x + 5y – 14 = 0

Answered by Pragya Singh | 11 months ago

Related Questions

An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Class 11 Maths Conic Sections View Answer

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Class 11 Maths Conic Sections View Answer

Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.

Class 11 Maths Conic Sections View Answer

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Class 11 Maths Conic Sections View Answer

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Class 11 Maths Conic Sections View Answer